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Question:

Three randomly chosen non-negative integers x, y and z are found to satisfy the equation x + y + z = 10. Then the probability that z is even, is

511

12

611

3655

Solution:

Assumezto be0,2,4,6,8,10and find solution ofx+y=10−z, we getCase 1:z=0⇒x+y=10andx,y∈0,1,2,...,10This can be done in10+2𕒵C2𕒵=11C1ways.Case 2:z=2⇒x+y=8andx,y∈0,1,2,,8This can be done in8+2𕒵C2𕒵=9C1ways.Similarly, we can obtain for all the cases as7C1,5C1,3C1and1C1ways.The total number of solutions for the given expression is10+3𕒵C3𕒵=12C2ways.Hence, probability=11C1+9C1+7C1+5C1+3C1+1C112C2=3666=611Hence, option C is correct.